Biostat 202B Homework 2

Due April 18, 2024 @ 11:59PM

Question 1

Answer:

\[\begin{align*} nS_{n}^2 &= \sum_{i=1}^{n} (X_i-\bar{X})^2 \\ &= \sum_{i=1}^{n} ((X_i-\mu)-(\bar{X}-\mu))^2 \\ &= \sum_{i=1}^{n} ((X_i-\mu)^2 - 2(X_i-\mu)(\bar{X}-\mu) + (\bar{X}-\mu)^2) \\ &= \sum_{i=1}^{n} (X_i-\mu)^2 - 2(\bar{X}-\mu)\sum_{i=1}^{n} (X_i-\mu) + n(\bar{X}-\mu)^2 \\ &= \sum_{i=1}^{n} (X_i-\mu)^2 - 2n(\bar{X}-\mu)^2 + n(\bar{X}-\mu)^2 \\ &= \sum_{i=1}^{n} (X_i-\mu)^2 - n(\bar{X}-\mu)^2 \end{align*}\]

Hence, \(S_n^2=\frac{1}{n}\sum_{i=1}^{n} (X_i-\mu)^2 - (\bar{X}-\mu)^2\)

Let \(Y_i=(X_i-\mu)^2\), then \(E(Y_i)=\mu_y=\sigma^2\)

\(Var(Y_i)=E(X_i-\mu)^4-\{E(X_i-\mu)^2\}^2=E(X_i-\mu_x)^4-\sigma^4\)

Since \(\gamma = \frac{E(X_i-\mu_x)^4}{\sigma^4}\)

\(Var(Y_i)=\sigma^4(\gamma-1)\)

\(S_n^2=\frac{1}{n}\sum_{i=1}^{n} Y_i - (\bar{X}-\mu)^2=\bar{Y}-(\bar{X}-\mu)^2\)

Consider \(\sqrt{n}(S_n^2-\sigma_x^2)=\sqrt{n}\{\bar{Y}-(\bar{X}-\mu)^2-\sigma_x^2\}=\sqrt{n}\{\bar{Y}-(\bar{X}-\mu)^2-\mu_y\}=\sqrt{n}(\bar{Y}-\mu_y)-\sqrt{n}(\bar{X}-\mu)^2\)

By CLT, \(\sqrt{n}(\bar{Y}-\mu_y)\overset{D}{\rightarrow}\sigma_yZ\sim N(0,\sigma_y^2)\)

By CLT, \(\sqrt{n}(\bar{X}-\mu)\overset{D}{\rightarrow}N(0,\sigma^2)\)

By slutsky, \(\bar{X}-\mu = \frac{1}{\sqrt{n}}\sqrt{n}(\bar{X}-\mu)\overset{P}{\rightarrow}0\)

By C.M.T, \(\sqrt{n}(S_n^2-\sigma_x^2)\overset{D}{\rightarrow}N(0,\sigma^4(\gamma-1))\)

Question 2

Answer:

Since \(\frac{1}{\sqrt{n}}\rightarrow 0\) as \(n\rightarrow \infty\), we have

\(Y_n-\theta \rightarrow 0\) as \(n\rightarrow \infty\).

\(Y_n\rightarrow \theta\) in probability. (Note: There is a typo in the question)

Question 3

Answer:

By WLLN, \(\bar{X} \overset{P}{\rightarrow} p\)

\(E(X)=p\) and \(Var(X)=p(1-p)\)

By CLT,

\(\frac{\sqrt{n}(\bar{X}-p)}{\sqrt{p(1-p)}} \overset{D}{\rightarrow} N(0,1)\)

Since \(\frac{p(1-p)}{\sqrt{\bar{X}(1-\bar{X})}}\overset{P}{\rightarrow}1\)

\(\frac{\sqrt{n}(\bar{X}-p)}{\sqrt{\bar{X}(1-\bar{X})}} \overset{D}{\rightarrow} N(0,1)\)

\(P(|\frac{\sqrt{n}(\bar{X}-p)}{\sqrt{\bar{X}(1-\bar{X})}}|\leq 1.96)=0.95\)

\(P(-1.96\leq \frac{\sqrt{n}(\bar{X}-p)}{\sqrt{\bar{X}(1-\bar{X})}}\leq 1.96)=0.95\)

\(P(\frac{-1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}}\leq \bar{X}-p\leq \frac{1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}})=0.95\)

\(P(\bar{X}-\frac{1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}}\leq p\leq \bar{X}+\frac{1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}})=0.95\)

Hence, 95% asymptotic CI for p is \((\bar{X}-\frac{1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}}, \bar{X}+\frac{1.96\sqrt{\bar{X}(1-\bar{X})}}{\sqrt{n}})\)

Question 4

Answer:

Let \(N=n+m\), \(\gamma\) is Kurtosis of X

By WLLN and C.M.T,

\(\sqrt{N}\sqrt{\frac{S_X^2}{n}+\frac{S_Y^2}{m}}=\sqrt{\frac{S_X^2}{n/N}+\frac{S_Y^2}{m/N}}\overset{P}{\rightarrow}\sqrt{\frac{\sigma_X^2}{\gamma}+\frac{\sigma_Y^2}{1-\gamma}}\)

\(\sqrt{N}\{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)\}=\sqrt{\frac{N}{n}}\sqrt{n}(\bar{X}-\mu_X)-\sqrt{\frac{N}{m}}\sqrt{m}(\bar{Y}-\mu_Y)\)

Then,

\[\begin{align*} M(t)&=E(e^{t\sqrt{\frac{N}{n}}\sqrt{n}(\bar{X}-\mu_X)-t\sqrt{\frac{N}{m}}\sqrt{m}(\bar{Y}-\mu_Y)})\\ &=E\{e^{t\sqrt{\frac{N}{n}}\sigma_X\frac{\sqrt{n}(\bar{X}-\mu_X)}{\sigma_X}}\}E\{e^{t\sqrt{\frac{N}{m}}\sigma_Y\frac{\sqrt{m}(\bar{Y}-\mu_Y)}{\sigma_Y}}\}\\ &=M_Z(t\sqrt{\frac{1}{\gamma}}\sigma_X)M_Z(t\sqrt{\frac{1}{1-\gamma}\sigma_Y})\\ &=e^{\frac{t^2}{2}(\frac{\sigma_X^2}{\gamma}+\frac{\sigma_Y^2}{1-\gamma})} \text{, }\forall t \end{align*}\]

Hence,

\(\sqrt{N}\{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)\}\overset{D}{\rightarrow}N(0,\frac{\sigma_X^2}{\gamma}+\frac{\sigma_Y^2}{1-\gamma})\)

By combining the above results,

\(\frac{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)}{\sqrt{\frac{S_X^2}{n}+\frac{S_Y^2}{m}}}\overset{D}{\rightarrow}Z\sim N(0,1)\)

Question 5

(a) Answer:

Since \(X_i \sim Poisson(\theta)\),

BY WLLN,

\(\frac{\sqrt{n}(\bar{X}-\theta)}{\sqrt{\theta}} \overset{D}{\rightarrow} N(0,1)\)

\(\sqrt{n}\{\bar{X}-\theta\} \overset{D}{\rightarrow} N(0,\theta)\)

(b) Answer:

By delta method,

\(\sqrt{n}\{g(\bar{X})-g(\theta)\} \overset{D}{\rightarrow} N(0,\theta^2(g'(\theta))^2)\)

Question 6

(a) Answer:

To ensure \([g'(\theta)]^2\sigma^2(\theta)=1\), we need \(g'(\theta)=\frac{1}{\sqrt{\sigma^2}}\)

Since \(X_1,X_2,...,X_n \overset{\text{i.i.d.}}{\sim} Exponential(\frac{1}{\theta})\)

Then, \(\sigma^2=\theta^2\)

\(g'(\theta)=\frac{1}{\sqrt{\theta^2}}=\frac{1}{\theta}\)

\(g(\theta)=\ln(\theta)\)

Hence, \(g(X_n)=\ln(X_n)\) is the variance stabilizing transformation.

(b) Answer:

Similarly, we need \(g'(\theta)=\frac{1}{\sqrt{\sigma^2}}\)

Since \(X_1,X_2,...,X_n \overset{\text{i.i.d.}}{\sim} Poisson(\theta)\)

Then, \(\sigma^2=\theta\)

\(g'(\theta)=\frac{1}{\sqrt{\theta}}\)

\(g(\theta)=2\sqrt{\theta}\)

Hence, \(g(X_n)=2\sqrt{X_n}\) is the variance stabilizing transformation.

(c) Answer:

Similarly, we need \(g'(\theta)=\frac{1}{\sqrt{\sigma^2}}\)

Since \(X_1,X_2,...,X_n \overset{\text{i.i.d.}}{\sim} Bernoulli(\theta)\)

Then, \(\sigma^2=\theta(1-\theta)\)

\(g'(\theta)=\frac{1}{\sqrt{\theta(1-\theta)}}\)

\(g(\theta)=\arcsin(2\theta-1)\)

Hence, \(g(X_n)=\arcsin(2X_n-1)\) is the variance stabilizing transformation.

(d) Answer:

\(\sqrt{n}(X_n-\theta) \overset{D}{\rightarrow} N(0,\sigma^2)\)

After transformation, we have

\(\sqrt{n}(2\sqrt{X_n}-2\sqrt{\theta}) \overset{D}{\rightarrow} N(0,1)\)

\(P(-1.96\leq \sqrt{n}(2\sqrt{X_n}-2\sqrt{\theta})\leq 1.96)=0.95\)

\(P(\frac{-1.96}{\sqrt{n}}\leq 2\sqrt{X_n}-2\sqrt{\theta}\leq\frac{1.96}{\sqrt{n}})=0.95\)

\(P(\frac{-1.96}{2\sqrt{n}}\leq \sqrt{X_n}-\sqrt{\theta}\leq\frac{1.96}{2\sqrt{n}})=0.95\)

\(P(\sqrt{X_n}-\frac{1.96}{2\sqrt{n}}\leq \sqrt{\theta}\leq\sqrt{X_n}+\frac{1.96}{2\sqrt{n}})=0.95\)

\(P((\sqrt{X_n}-\frac{1.96}{2\sqrt{n}})^2\leq \theta\leq(\sqrt{X_n}+\frac{1.96}{2\sqrt{n}})^2)=0.95\)